MOTION IN A STRAIGHT LINE

CHAPTER THREE MOTION IN A STRAIGHT LINE 3. Introduction 3. Position, path length and displacement 3.3 Average velocity and average speed 3.4 Instantaneous velocity and speed 3.5 Acceleration 3.6 Kinematic equations for uniformly accelerated motion 3.7 Relative velocity Summary Points to ponder Eercises Additional eercises Appendi 3. 3. INTRODUCTION Motion is common to everything in the universe. We walk, run and ride a bicycle. Even when we are sleeping, air moves into and out of our lungs and blood flows in arteries and veins. We see leaves falling from trees and water flowing down a dam. Automobiles and planes carry people from one place to the other. The earth rotates once every twenty-four hours and revolves round the sun once in a year. The sun itself is in motion in the Milky Way, which is again moving within its local group of galaies. Motion is change in position of an object with time. How does the position change with time? In this chapter, we shall learn how to describe motion. For this, we develop the concepts of velocity and acceleration. We shall confine ourselves to the study of motion of objects along a straight line, also known as rectilinear motion. For the case of rectilinear motion with uniform acceleration, a set of simple equations can be obtained. Finally, to understand the relative nature of motion, we introduce the concept of relative velocity. In our discussions, we shall treat the objects in motion as point objects. This approimation is valid so far as the size of the object is much smaller than the distance it moves in a reasonable duration of time. In a good number of situations in real-life, the size of objects can be neglected and they can be considered as point-like objects without much error. In Kinematics, we study ways to describe motion without going into the causes of motion. What causes motion described in this chapter and the net chapter forms the subject matter of Chapter 5. 3. POSITION, PATH LENGTH AND DISPLACEMENT Earlier you learnt that motion is change in position of an object with time. In order to specify position, we need to use a reference point and a set of aes. It is convenient to choose

40 a rectangular coordinate system consisting of three mutually perpenducular aes, labelled X-, Y-, and Z- aes. The point of intersection of these three aes is called origin (O) and serves as the reference point. The coordinates (, y. z) of an object describe the position of the object with respect to this coordinate system. To measure time, we position a clock in this system. This coordinate system along with a clock constitutes a frame of reference. If one or more coordinates of an object change with time, we say that the object is in motion. Otherwise, the object is said to be at rest with respect to this frame of reference. The choice of a set of aes in a frame of reference depends upon the situation. For eample, for describing motion in one dimension, we need only one ais. To describe motion in two/three dimensions, we need a set of two/ three aes. Description of an event depends on the frame of reference chosen for the description. For eample, when you say that a car is moving on a road, you are describing the car with respect to a frame of reference attached to you or to the ground. But with respect to a frame of reference attached with a person sitting in the car, the car is at rest. To describe motion along a straight line, we can choose an ais, say X-ais, so that it coincides with the path of the object. We then measure the position of the object with reference to a conveniently chosen origin, say O, as shown in Fig. 3.. Positions to the right of O are taken as positive and to the left of O, as negative. Following this convention, the position coordinates of point P and Q in Fig. 3. are +360 m and +40 m. Similarly, the position coordinate of point R is 0 m. Path length Consider the motion of a car along a straight line. We choose the -ais such that it coincides PHYSICS with the path of the car s motion and origin of the ais as the point from where the car started moving, i.e. the car was at = 0 at t = 0 (Fig.3.). Let P, Q and R represent the positions of the car at different instants of time. Consider two cases of motion. In the first case, the car moves from O to P. Then the distance moved by the car is OP = +360 m. This distance is called the path length traversed by the car. In the second case, the car moves from O to P and then moves back from P to Q. During this course of motion, the path length traversed is OP + PQ = + 360 m + (+0 m) = + 480 m. Path length is a scalar quantity a quantity that has a magnitude only and no direction (see Chapter 4). Displacement It is useful to define another quantity displacement as the change in position. Let and be the positions of an object at time t and t. Then its displacement, denoted by, in time t = (t - t ), is given by the difference between the final and initial positions : = (We use the Greek letter delta ( ) to denote a change in a quantity.) If >, is positive; and if <, is negative. Displacement has both magnitude and direction. Such quantities are represented by vectors. You will read about vectors in the net chapter. Presently, we are dealing with motion along a straight line (also called rectilinear motion) only. In one-dimensional motion, there are only two directions (backward and forward, upward and downward) in which an object can move, and these two directions can easily be specified by + and signs. For eample, displacement of the car in moving from O to P is : = = (+360 m) 0 m = +360 m The displacement has a magnitude of 360 m and is directed in the positive direction as indicated by the + sign. Similarly, the displacement of the car from P to Q is 40 m 360 m = 0 m. The Fig. 3. -ais, origin and positions of a car at different times.

MOTION IN A STRAIGHT LINE 4 negative sign indicates the direction of displacement. Thus, it is not necessary to use vector notation for discussing motion of objects in one-dimension. The magnitude of displacement may or may not be equal to the path length traversed by an object. For eample, for motion of the car from O to P, the path length is +360 m and the displacement is +360 m. In this case, the magnitude of displacement (360 m) is equal to the path length (360 m). But consider the motion of the car from O to P and back to Q. In this case, the path length = (+360 m) + (+0 m) = + 480 m. However, the displacement = (+40 m) (0 m) = + 40 m. Thus, the magnitude of displacement (40 m) is not equal to the path length (480 m). The magnitude of the displacement for a course of motion may be zero but the corresponding path length is not zero. For eample, if the car starts from O, goes to P and then returns to O, the final position coincides with the initial position and the displacement is zero. However, the path length of this journey is OP + PO = 360 m + 360 m = 70 m. Motion of an object can be represented by a position-time graph as you have already learnt about it. Such a graph is a powerful tool to represent and analyse different aspects of motion of an object. For motion along a straight line, say X-ais, only -coordinate varies with time and we have an -t graph. Let us first consider the simple case in which an object is stationary, e.g. a car standing still at = 40 m. The position-time graph is a straight line parallel to the time ais, as shown in Fig. 3.(a). If an object moving along the straight line covers equal distances in equal intervals of time, it is said to be in uniform motion along a straight line. Fig. 3.(b) shows the position-time graph of such a motion. Fig. 3. Position-time graph of (a) stationary object, and (b) an object in uniform motion. (m) Fig. 3.3 Position-time graph of a car. t (s)

4 Now, let us consider the motion of a car that starts from rest at time t = 0 s from the origin O and picks up speed till t = 0 s and thereafter moves with uniform speed till t = 8 s. Then the brakes are applied and the car stops at t = 0 s and = 96 m. The position-time graph for this case is shown in Fig. 3.3. We shall refer to this graph in our discussion in the following sections. 3.3 AVERAGE VELOCITY AND AVERAGE SPEED When an object is in motion, its position changes with time. But how fast is the position changing with time and in what direction? To describe this, we define the quantity average velocity. Average velocity is defined as the change in position or displacement ( ) divided by the time intervals ( t), in which the displacement occurs : v = t t = t (3.) where and are the positions of the object at time t and t, respectively. Here the bar over the symbol for velocity is a standard notation used to indicate an average quantity. The SI unit for velocity is m/s or m s, although km h is used in many everyday applications. Like displacement, average velocity is also a vector quantity. But as eplained earlier, for motion in a straight line, the directional aspect of the vector can be taken care of by + and signs and we do not have to use the vector notation for velocity in this chapter. Fig. 3.4 The average velocity is the slope of line P P. PHYSICS Consider the motion of the car in Fig. 3.3. The portion of the -t graph between t = 0 s and t = 8 s is blown up and shown in Fig. 3.4. As seen from the plot, the average velocity of the car between time t = 5 s and t = 7 s is : ( 7. 4 00. ) m v = = = 8.7 m s t t ( 7 5) s Geometrically, this is the slope of the straight line P P connecting the initial position P to the final position P as shown in Fig. 3.4. The average velocity can be positive or negative depending upon the sign of the displacement. It is zero if the displacement is zero. Fig. 3.5 shows the -t graphs for an object, moving with positive velocity (Fig. 3.5a), moving with negative velocity (Fig. 3.5b) and at rest (Fig. 3.5c). Fig. 3.5 Position-time graph for an object (a) moving with positive velocity, (b) moving with negative velocity, and (c) at rest. Average velocity as defined above involves only the displacement of the object. We have seen earlier that the magnitude of displacement may be different from the actual path length. To describe the rate of motion over the actual path, we introduce another quantity called average speed. Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place : Total path length Average speed = (3.) Total time interval Average speed has obviously the same unit (m s ) as that of velocity. But it does not tell us in what direction an object is moving. Thus, it is always positive (in contrast to the average velocity which can be positive or negative). If the motion of an object is along a straight line and in the same direction, the magnitude of displacement is equal to the total path length. In that case, the magnitude of average velocity

MOTION IN A STRAIGHT LINE 43 is equal to the average speed. This is not always the case, as you will see in the following eample. Eample 3. A car is moving along a straight line, say OP in Fig. 3.. It moves from O to P in 8 s and returns from P to Q in 6.0 s. What are the average velocity and average speed of the car in going (a) from O to P? and (b) from O to P and back to Q? 3.4 INSTANTANEOUS VELOCITY AND SPEED The average velocity tells us how fast an object has been moving over a given time interval but does not tell us how fast it moves at different instants of time during that interval. For this, we define instantaneous velocity or simply velocity v at an instant t. The velocity at an instant is defined as the limit of the average velocity as the time interval t becomes infinitesimally small. In other words, Answer (a) Displacement Average velocity = Time interval Average speed + 360 m v = = + 0 m s 8 s = Path length Time interval 360 m = = 0 m s 8 s Thus, in this case the average speed is equal to the magnitude of the average velocity. (b) In this case, Displacement + 40 m Average velocity = = Time interval 8 6.0 s Path length Average speed = = Time interval ( + ) - =+0 ms OP + PQ t ( 360 +0 ) m - = = 0 m s 4 s Thus, in this case the average speed is not equal to the magnitude of the average velocity. This happens because the motion here involves change in direction so that the path length is greater than the magnitude of displacement. This shows that speed is, in general, greater than the magnitude of the velocity. If the car in Eample 3. moves from O to P and comes back to O in the same time interval, average speed is 0 m/s but the average velocity is zero! v = = d dt lim t 0 t lim t 0 (3.3a) (3.3b) where the symbol stands for the operation of taking limit as t 0 of the quantity on its right. In the language of calculus, the quantity on the right hand side of Eq. (3.3a) is the differential coefficient of with respect to t and is denoted by d (see Appendi 3.). It is the dt rate of change of position with respect to time, at that instant. We can use Eq. (3.3a) for obtaining the value of velocity at an instant either graphically or numerically. Suppose that we want to obtain graphically the value of velocity at time t = 4 s (point P) for the motion of the car represented in Fig. 3.3. The figure has been redrawn in Fig.3.6 choosing different scales to facilitate the Fig. 3.6 Determining velocity from position-time graph. Velocity at t = 4 s is the slope of the tangent to the graph at that instant.

44 calculation. Let us take t = s centred at t = 4 s. Then, by the definition of the average velocity, the slope of line P P ( Fig. 3.6) gives the value of average velocity over the interval 3 s to 5 s. Now, we decrease the value of t from s to s. Then line P P becomes Q Q and its slope gives the value of the average velocity over the interval 3.5 s to 4.5 s. In the limit t 0, the line P P becomes tangent to the positiontime curve at the point P and the velocity at t = 4 s is given by the slope of the tangent at that point. It is difficult to show this process graphically. But if we use numerical method to obtain the value of the velocity, the meaning of the limiting process becomes clear. For the graph shown in Fig. 3.6, = 0.08 t 3. Table 3. gives the value of / t calculated for t equal to.0 s,.0 s, 0.5 s, 0. s and 0.0 s centred at t = 4.0 s. The second and third columns give the value of t = t t t and t t = + and the fourth and the fifth columns give the corresponding values of, i.e. (t ) = 0.08 t 3 and (t ) = 0.08 t 3. The sith column lists the difference = (t ) (t ) and the last column gives the ratio of and t, i.e. the average velocity corresponding to the value of t listed in the first column. We see from Table 3. that as we decrease the value of t from.0 s to 0.00 s, the value of the average velocity approaches the limiting value 3.84 m s which is the value of velocity at t = 4.0 s, i.e. the value of d at t = 4.0 s. In this dt manner, we can calculate velocity at each PHYSICS instant for motion of the car shown in Fig. 3.3. For this case, the variation of velocity with time is found to be as shown in Fig. 3.7. Fig. 3.7 Velocity time graph corresponding to motion shown in Fig. 3.3. The graphical method for the determination of the instantaneous velocity is always not a convenient method. For this, we must carefully plot the position time graph and calculate the value of average velocity as t becomes smaller and smaller. It is easier to calculate the value of velocity at different instants if we have data of positions at different instants or eact epression for the position as a function of time. Then, we calculate / t from the data for decreasing the value of t and find the limiting value as we have done in Table 3. or use differential calculus for the given epression and calculate d at different instants as done in dt the following eample. Table 3. Limiting value of t at t = 4 s

MOTION IN A STRAIGHT LINE 45 Eample 3. The position of an object moving along -ais is given by = a + bt where a = 8.5 m, b =.5 m s and t is measured in seconds. What is its velocity at t = 0 s and t =.0 s. What is the average velocity between t =.0 s and t = 4.0 s? Answer In notation of differential calculus, the velocity is ( ) v = = dt d dt a + bt = b t = 5.0 t m s - At t = 0 s, v = 0 m s and at t =.0 s, v = 0 m s -. ( 4.0) (.0) Average velocity = 4.0.0 a + 6 b a 4b = = 6.0 b.0 - = 6.0.5 =5 m s From Fig. 3.7, we note that during the period t =0 s to 8 s the velocity is constant. Between period t =8 s to t = 0 s, it is uniformly decreasing and during the period t = 0 s to t = 0 s, it is increasing. Note that for uniform motion, velocity is the same as the average velocity at all instants. Instantaneous speed or simply speed is the magnitude of velocity. For eample, a velocity of + 4.0 m s and a velocity of 4.0 m s both have an associated speed of 4.0 m s -. It should be noted that though average speed over a finite interval of time is greater or equal to the magnitude of the average velocity, instantaneous speed at an instant is equal to the magnitude of the instantaneous velocity at that instant. Why so? 3.5 ACCELERATION The velocity of an object, in general, changes during its course of motion. How to describe this change? Should it be described as the rate of change in velocity with distance or with time? This was a problem even in Galileo s time. It was first thought that this change could be described by the rate of change of velocity with distance. But, through his studies of motion of freely falling objects and motion of objects on an inclined plane, Galileo concluded that the rate of change of velocity with time is a constant of motion for all objects in free fall. On the other hand, the change in velocity with distance is not constant it decreases with the increasing distance of fall. This led to the concept of acceleration as the rate of change of velocity with time. The average acceleration a over a time interval is defined as the change of velocity divided by the time interval : a v v v = = t t t (3.4) where v and v are the instantaneous velocities or simply velocities at time t and t. It is the average change of velocity per unit time. The SI unit of acceleration is m s. On a plot of velocity versus time, the average acceleration is the slope of the straight line connecting the points corresponding to (v, t ) and (v, t ). The average acceleration for velocity-time graph shown in Fig. 3.7 for different time intervals 0 s - 0 s, 0 s 8 s, and 8 s 0 s are : 0 s - 0 s 0 s - 8 s 8 s - 0 s a (m s ) ( ) ( ) 4 0 m s a = =.4 m s 0 0 s ( ) ( ) 4 4 m s a = = 0 m s 8 0 s ( ) ( ) 0 4 m s a = = m s 0 8 s Fig. 3.8 Acceleration as a function of time for motion represented in Fig. 3.3. Instantaneous acceleration is defined in the same way as the instantaneous velocity : v dv a = lim = (3.5) t 0 t d t The acceleration at an instant is the slope of the tangent to the v t curve at that instant. For the v t curve shown in Fig. 3.7, we can obtain acceleration at every instant of time. The resulting a t curve is shown in Fig. 3.8. We see

46 that the acceleration is nonuniform over the period 0 s to 0 s. It is zero between 0 s and 8 s and is constant with value m s between 8 s and 0 s. When the acceleration is uniform, obviously, it equals the average acceleration over that period. Since velocity is a quantity having both magnitude and direction, a change in velocity may involve either or both of these factors. Acceleration, therefore, may result from a change in speed (magnitude), a change in direction or changes in both. Like velocity, acceleration can also be positive, negative or zero. Position-time graphs for motion with positive, negative and zero acceleration are shown in Figs. 3.9 (a), (b) and (c), respectively. Note that the graph curves upward for positive acceleration; downward for negative acceleration and it is a straight line for zero acceleration. As an eercise, identify in Fig. 3.3, the regions of the curve that correspond to these three cases. Although acceleration can vary with time, our study in this chapter will be restricted to motion with constant acceleration. In this case, the average acceleration equals the constant value of acceleration during the interval. If the velocity of an object is v o at t = 0 and v at time t, we have v v0 a = or, v = v0 + a t t 0 (3.6) PHYSICS (b) An object is moving in positive direction with a negative acceleration, for eample, motion of the car in Fig 3.3 between t = 8 s and 0 s. (c) An object is moving in negative direction with a negative acceleration, for eample the motion of a car moving from O in Fig. 3. in negative -direction with increasing speed. (d) An object is moving in positive direction till time t, and then turns back with the same negative acceleration, for eample the motion of a car from point O to point Q in Fig. 3. till time t with decreasing speed and turning back and moving with the same negative acceleration. An interesting feature of a velocity-time graph for any moving object is that the area under the curve represents the displacement over a given time interval. A general proof of this Fig. 3.9 Position-time graph for motion with (a) positive acceleration; (b) negative acceleration, and (c) zero acceleration. Let us see how velocity-time graph looks like for some simple cases. Fig. 3.0 shows velocitytime graph for motion with constant acceleration for the following cases : (a) An object is moving in a positive direction with a positive acceleration, for eample the motion of the car in Fig. 3.3 between t = 0 s and t = 0 s. Fig. 3.0 Velocity time graph for motions with constant acceleration. (a) Motion in positive direction with positive acceleration, (b) Motion in positive direction with negative acceleration, (c) Motion in negative direction with negative acceleration, (d) Motion of an object with negative acceleration that changes direction at time t. Between times 0 to t, its moves in positive - direction and between t and t it moves in the opposite direction.

MOTION IN A STRAIGHT LINE 47 statement requires use of calculus. We can, however, see that it is true for the simple case of an object moving with constant velocity u. Its velocity-time graph is as shown in Fig. 3.. = ( v v0 ) t + v0t Fig. 3. Area under v t curve equals displacement of the object over a given time interval. The v-t curve is a straight line parallel to the time ais and the area under it between t = 0 and t = T is the area of the rectangle of height u and base T. Therefore, area = u T = ut which is the displacement in this time interval. How come in this case an area is equal to a distance? Think! Note the dimensions of quantities on the two coordinate aes, and you will arrive at the answer. Note that the -t, v-t, and a-t graphs shown in several figures in this chapter have sharp kinks at some points implying that the functions are not differentiable at these points. In any realistic situation, the functions will be differentiable at all points and the graphs will be smooth. What this means physically is that acceleration and velocity cannot change values abruptly at an instant. Changes are always continuous. 3.6 KINEMATIC EQUATIONS FOR UNIFORMLY ACCELERATED MOTION For uniformly accelerated motion, we can derive some simple equations that relate displacement (), time taken (t), initial velocity (v 0 ), final velocity (v) and acceleration (a). Equation (3.6) already obtained gives a relation between final and initial velocities v and v 0 of an object moving with uniform acceleration a : v = v 0 + at (3.6) This relation is graphically represented in Fig. 3.. The area under this curve is : Area between instants 0 and t = Area of triangle ABC + Area of rectangle OACD Fig. 3. Area under v-t curve for an object with uniform acceleration. As eplained in the previous section, the area under v-t curve represents the displacement. Therefore, the displacement of the object is : = ( v v0 ) t + v0t (3.7) But v v0 = a t Therefore, = a t + v0t or, v t at Equation (3.7) can also be written as where, v = 0 + (3.8) v + v 0 = t = v t (3.9a) v + v 0 = (constant acceleration only) (3.9b) Equations (3.9a) and (3.9b) mean that the object has undergone displacement with an average velocity equal to the arithmetic average of the initial and final velocities. From Eq. (3.6), t = (v v 0 )/a. Substituting this in Eq. (3.9a), we get v + v v v v v = v t = = a a 0 0 0 0 v = v + a (3.0)

48 This equation can also be obtained by substituting the value of t from Eq. (3.6) into Eq. (3.8). Thus, we have obtained three important equations : v = v0 + at = v0t + at 0 v = v + a (3.a) connecting five quantities v 0, v, a, t and. These are kinematic equations of rectilinear motion for constant acceleration. The set of Eq. (3.a) were obtained by assuming that at t = 0, the position of the particle, is 0. We can obtain a more general equation if we take the position coordinate at t = 0 as non-zero, say 0. Then Eqs. (3.a) are modified (replacing by 0 ) to : v = v0 + at v t at = 0 + 0 + (3.b) v = v + a( ) (3.c) 0 0 We can write t 0 ( ) = v + at dt 0 = v0 t + a t = 0 0 + v0 t + a t d d d d a = v = v = v v dt dt or, v dv = a Integrating both sides, v v v dv v0 0 v 0 = = a ( ) a 0 ( ) v = v + a 0 0 PHYSICS The advantage of this method is that it can be used for motion with non-uniform acceleration also. Now, we shall use these equations to some important cases. Eample 3.3 Obtain equations of motion for constant acceleration using method of calculus. Answer By definition dv a = dt dv = a dt Integrating both sides v dv = a dt v0 0 v v = at 0 t t d 0 = a t (a is constant) v = v0 + at Further, v = dt = v dt Integrating both sides 0 = t v 0 d t Eample 3.4 A ball is thrown vertically upwards with a velocity of 0 m s from the top of a multistorey building. The height of the point from where the ball is thrown is 5.0 m from the ground. (a) How high will the ball rise? and (b) how long will it be before the ball hits the ground? Take g = 0 m s. Answer (a) Let us take the y-ais in the vertically upward direction with zero at the ground, as shown in Fig. 3.3. Now v o = + 0 m s, a = g = 0 m s, v = 0 m s If the ball rises to height y from the point of launch, then using the equation v = v0 + a ( y y0 ) we get 0 = (0) + ( 0)(y y 0 ) Solving, we get, (y y 0 ) = 0 m. (b) We can solve this part of the problem in two ways. Note carefully the methods used.

MOTION IN A STRAIGHT LINE 49 0 = 5 +0 t + (½) (-0) t Or, 5t 0t 5 = 0 Solving this quadratic equation for t, we get t = 5s Note that the second method is better since we do not have to worry about the path of the motion as the motion is under constant acceleration. Eample 3.5 Free-fall : Discuss the motion of an object under free fall. Neglect air resistance. Fig. 3.3 FIRST METHOD : In the first method, we split the path in two parts : the upward motion (A to B) and the downward motion (B to C) and calculate the corresponding time taken t and t. Since the velocity at B is zero, we have : v = v o + at 0 = 0 0t Or, t = s This is the time in going from A to B. From B, or the point of the maimum height, the ball falls freely under the acceleration due to gravity. The ball is moving in negative y direction. We use equation y = y0 + v0t + at We have, y 0 = 45 m, y = 0, v 0 = 0, a = g = 0 m s 0 = 45 + (½) ( 0) t Solving, we get t = 3 s Therefore, the total time taken by the ball before it hits the ground = t + t = s + 3 s = 5 s. SECOND METHOD : The total time taken can also be calculated by noting the coordinates of initial and final positions of the ball with respect to the origin chosen and using equation y = y0 + v0t + at Now y 0 = 5 m y = 0 m v o = 0 m s -, a = 0m s, t =? Answer An object released near the surface of the Earth is accelerated downward under the influence of the force of gravity. The magnitude of acceleration due to gravity is represented by g. If air resistance is neglected, the object is said to be in free fall. If the height through which the object falls is small compared to the earth s radius, g can be taken to be constant, equal to 9.8 m s. Free fall is thus a case of motion with uniform acceleration. We assume that the motion is in y-direction, more correctly in y-direction because we choose upward direction as positive. Since the acceleration due to gravity is always downward, it is in the negative direction and we have a = g = 9.8 m s The object is released from rest at y = 0. Therefore, v 0 = 0 and the equations of motion become: v = 0 g t = 9.8 t m s y = 0 ½ g t = 4.9 t m v = 0 g y = 9.6 y m s These equations give the velocity and the distance travelled as a function of time and also the variation of velocity with distance. The variation of acceleration, velocity, and distance, with time have been plotted in Fig. 3.4(a), (b) and (c). (a)

50 PHYSICS (b) traversed during successive intervals of time. Since initial velocity is zero, we have y = gt Using this equation, we can calculate the position of the object after different time intervals, 0, τ, τ, 3τ which are given in second column of Table 3.. If we take ( /) gτ as y 0 the position coordinate after first time interval, then third column gives the positions in the unit of y o. The fourth column gives the distances traversed in successive s. We find that the distances are in the simple ratio : 3: 5: 7: 9: as shown in the last column. This law was established by Galileo Galilei (564-64) who was the first to make quantitative studies of free fall. Fig. 3.4 Motion of an object under free fall. (a) Variation of acceleration with time. (b) Variation of velocity with time. (c) Variation of distance with time (c) Eample 3.6 Galileo s law of odd numbers : The distances traversed, during equal intervals of time, by a body falling from rest, stand to one another in the same ratio as the odd numbers beginning with unity [namely, : 3: 5: 7...]. Prove it. Eample 3.7 Stopping distance of vehicles : When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v 0 ) and the braking capacity, or deceleration, a that is caused by the braking. Derive an epression for stopping distance of a vehicle in terms of v o and a. Answer Let the distance travelled by the vehicle before it stops be d s. Then, using equation of motion v = v o + a, and noting that v = 0, we have the stopping distance Answer Let us divide the time interval of motion of an object under free fall into many equal intervals τ and find out the distances Table 3. Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the

MOTION IN A STRAIGHT LINE 5 initial velocity increases the stopping distance by a factor of 4 (for the same deceleration). For the car of a particular make, the braking distance was found to be 0 m, 0 m, 34 m and 50 m corresponding to velocities of, 5, 0 and 5 m/s which are nearly consistent with the above formula. Stopping distance is an important factor considered in setting speed limits, for eample, in school zones. Eample 3.8 Reaction time : When a situation demands our immediate action, it takes some time before we really respond. Reaction time is the time a person takes to observe, think and act. For eample, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time. Reaction time depends on compleity of the situation and on an individual. You can measure your reaction time by a simple eperiment. Take a ruler and ask your friend to drop it vertically through the gap between your thumb and forefinger (Fig. 3.5). After you catch it, find the distance d travelled by the ruler. In a particular case, d was found to be.0 cm. Estimate reaction time. Fig. 3.5 Measuring the reaction time. Answer The ruler drops under free fall. Therefore, v o = 0, and a = g = 9.8 m s. The distance travelled d and the reaction time t r are related by d = gt r d Or, t r = s g Given d =.0 cm and g = 9.8 m s the reaction time is 0. t r = s 0. s. 9.8 3.7 RELATIVE VELOCITY You must be familiar with the eperience of travelling in a train and being overtaken by another train moving in the same direction as you are. While that train must be travelling faster than you to be able to pass you, it does seem slower to you than it would be to someone standing on the ground and watching both the trains. In case both the trains have the same velocity with respect to the ground, then to you the other train would seem to be not moving at all. To understand such observations, we now introduce the concept of relative velocity. Consider two objects A and B moving uniformly with average velocities v A and v B in one dimension, say along -ais. (Unless otherwise specified, the velocities mentioned in this chapter are measured with reference to the ground). If A (0) and B (0) are positions of objects A and B, respectively at time t = 0, their positions A (t) and B (t) at time t are given by: A (t ) = A (0) + v A t (3.a) B (t) = B (0) + v B t (3.b) Then, the displacement from object A to object B is given by BA (t) = B (t) A (t) = [ B (0) A (0) ] + (v B v A ) t. (3.3) Equation (3.3) is easily interpreted. It tells us that as seen from object A, object B has a velocity v B v A because the displacement from A to B changes steadily by the amount v B v A in each unit of time. We say that the velocity of object B relative to object A is v B v A : v BA = v B v A (3.4a) Similarly, velocity of object A relative to object B is: v AB = v A v B (3.4b)

5 PHYSICS Fig. 3.6 Position-time graphs of two objects with equal velocities. This shows: v BA = v AB (3.4c) Now we consider some special cases : (a) If v B = v A, v B v A = 0. Then, from Eq. (3.3), B (t) A (t) = B (0) A (0). Therefore, the two objects stay at a constant distance ( B (0) A (0)) apart, and their position time graphs are straight lines parallel to each other as shown in Fig. 3.6. The relative velocity v AB or v BA is zero in this case. (b) If v A > v B, v B v A is negative. One graph is steeper than the other and they meet at a common point. For eample, suppose v A = 0 m s - and A (0) = 0 m; and v B = 0 m s -, B (0) = 40 m; then the time at which they meet is t = 3 s (Fig. 3.7). At this instant they are both at a position A (t) = B (t) = 70 m. Thus, object A overtakes object B at this time. In this case,v BA = 0 m s 0 m s = 0 m s = v AB. (c) Suppose v A and v B are of opposite signs. For eample, if in the above eample object A is moving with 0 m s starting at A (0) = 0 m and object B is moving with 0 m s starting at B (0) = 40 m, the two objects meet at t = s (Fig. 3.8). The velocity of B relative to A, v BA = [ 0 (0)] m s = 30 m s = v AB. In this case, the magnitude of v BA or v AB ( = 30 m s ) is greater than the magnitude of velocity of A or that of B. If the objects under consideration are two trains, then for a person sitting on either of the two, the other train seems to go very fast. Note that Eq. (3.4) are valid even if v A and v B represent instantaneous velocities. Fig. 3.7 Position-time graphs of two objects with unequal velocities, showing the time of meeting. Fig. 3.8 Position-time graphs of two objects with velocities in opposite directions, showing the time of meeting. t(s) Eample 3.9 Two parallel rail tracks run north-south. Train A moves north with a speed of 54 km h, and train B moves south with a speed of 90 km h. What is the (a) velocity of B with respect to A?, (b) velocity of ground with respect to B?, and (c) velocity of a monkey running on the roof of the train A against its motion (with a velocity of 8 km h with respect to the train A) as observed by a man standing on the ground? Answer Choose the positive direction of -ais to be from south to north. Then,

MOTION IN A STRAIGHT LINE 53 v A = + 54 km h = 5 m s v B = 90 km h = 5 m s Relative velocity of B with respect to A = v B v A = 40 m s, i.e. the train B appears to A to move with a speed of 40 m s from north to south. Relative velocity of ground with respect to B = 0 v B = 5 m s. In (c), let the velocity of the monkey with respect to ground be v M. Relative velocity of the monkey with respect to A, v MA = v M v A = 8 km h = 5 ms. Therefore, v M = (5 5) m s = 0 m s. SUMMARY. An object is said to be in motion if its position changes with time. The position of the object can be specified with reference to a conveniently chosen origin. For motion in a straight line, position to the right of the origin is taken as positive and to the left as negative.. Path length is defined as the total length of the path traversed by an object. 3. Displacement is the change in position : =. Path length is greater or equal to the magnitude of the displacement between the same points. 4. An object is said to be in uniform motion in a straight line if its displacement is equal in equal intervals of time. Otherwise, the motion is said to be non-uniform. 5. Average velocity is the displacement divided by the time interval in which the displacement occurs : v = t On an -t graph, the average velocity over a time interval is the slope of the line connecting the initial and final positions corresponding to that interval. 6. Average Speed is the ratio of total path length traversed and the corresponding time interval. The average speed of an object is greater or equal to the magnitude of the average velocity over a given time interval. 7. Instantaneous velocity or simply velocity is defined as the limit of the average velocity as the time interval t becomes infinitesimally small : v = lim v = lim = t 0 t 0 t d t The velocity at a particular instant is equal to the slope of the tangent drawn on position-time graph at that instant. 8. Average acceleration is the change in velocity divided by the time interval during which the change occurs : v a = t 9. Instantaneous acceleration is defined as the limit of the average acceleration as the time interval t goes to zero : v dv a = lim a = lim = t 0 t 0 t d t The acceleration of an object at a particular time is the slope of the velocity-time graph at that instant of time. For uniform motion, acceleration is zero and the -t graph is a straight line inclined to the time ais and the v-t graph is a straight line

54 PHYSICS parallel to the time ais. For motion with uniform acceleration, -t graph is a parabola while the v-t graph is a straight line inclined to the time ais. 0. The area under the velocity-time curve between times t and t is equal to the displacement of the object during that interval of time.. For objects in uniformly accelerated rectilinear motion, the five quantities, displacement, time taken t, initial velocity v 0, final velocity v and acceleration a are related by a set of simple equations called kinematic equations of motion : v = v 0 + at = v0t + at 0 v = v + a if the position of the object at time t = 0 is 0. If the particle starts at = 0, in above equations is replaced by ( 0 ).

MOTION IN A STRAIGHT LINE 55 POINTS TO PONDER. The path length traversed by an object between two points is, in general, not the same as the magnitude of displacement. The displacement depends only on the end points; the path length (as the name implies) depends on the actual path. In one dimension, the two quantities are equal only if the object does not change its direction during the course of motion. In all other cases, the path length is greater than the magnitude of displacement.. In view of point above, the average speed of an object is greater than or equal to the magnitude of the average velocity over a given time interval. The two are equal only if the path length is equal to the magnitude of displacement. 3. The origin and the positive direction of an ais are a matter of choice. You should first specify this choice before you assign signs to quantities like displacement, velocity and acceleration. 4. If a particle is speeding up, acceleration is in the direction of velocity; if its speed is decreasing, acceleration is in the direction opposite to that of the velocity. This statement is independent of the choice of the origin and the ais. 5. The sign of acceleration does not tell us whether the particle s speed is increasing or decreasing. The sign of acceleration (as mentioned in point 3) depends on the choice of the positive direction of the ais. For eample, if the vertically upward direction is chosen to be the positive direction of the ais, the acceleration due to gravity is negative. If a particle is falling under gravity, this acceleration, though negative, results in increase in speed. For a particle thrown upward, the same negative acceleration (of gravity) results in decrease in speed. 6. The zero velocity of a particle at any instant does not necessarily imply zero acceleration at that instant. A particle may be momentarily at rest and yet have non-zero acceleration. For eample, a particle thrown up has zero velocity at its uppermost point but the acceleration at that instant continues to be the acceleration due to gravity. 7. In the kinematic equations of motion [Eq. (3.)], the various quantities are algebraic, i.e. they may be positive or negative. The equations are applicable in all situations (for one dimensional motion with constant acceleration) provided the values of different quantities are substituted in the equations with proper signs. 8. The definitions of instantaneous velocity and acceleration (Eqs. (3.3) and (3.5)) are eact and are always correct while the kinematic equations (Eq. (3.)) are true only for motion in which the magnitude and the direction of acceleration are constant during the course of motion. EXERCISES 3. In which of the following eamples of motion, can the body be considered approimately a point object: (a) a railway carriage moving without jerks between two stations. (b) a monkey sitting on top of a man cycling smoothly on a circular track. (c) a spinning cricket ball that turns sharply on hitting the ground. (d) a tumbling beaker that has slipped off the edge of a table. 3. The position-time (-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.9. Choose the correct entries in the brackets below ; (a) (A/B) lives closer to the school than (B/A) (b) (A/B) starts from the school earlier than (B/A) (c) (A/B) walks faster than (B/A) (d) A and B reach home at the (same/different) time (e) (A/B) overtakes (B/A) on the road (once/twice).

56 PHYSICS Fig. 3.9 3.3 A woman starts from her home at 9.00 am, walks with a speed of 5 km h on a straight road up to her office.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 5 km h. Choose suitable scales and plot the -t graph of her motion. 3.4 A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is m long and requires s. Plot the -t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 3 m away from the start. 3.5 A jet airplane travelling at the speed of 500 km h ejects its products of combustion at the speed of 500 km h relative to the jet plane. What is the speed of the latter with respect to an observer on the ground? 3.6 A car moving along a straight highway with speed of 6 km h is brought to a stop within a distance of 00 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop? 3.7 Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 7 km h in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by m s. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them? 3.8 On a two-lane road, car A is travelling with a speed of 36 km h. Two cars B and C approach car A in opposite directions with a speed of 54 km h each. At a certain instant, when the distance AB is equal to AC, both being km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident? 3.9 Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 0 km h in the direction A to B notices that a bus goes past him every 8 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road? 3.0 A player throws a ball upwards with an initial speed of 9.4 m s. (a) What is the direction of acceleration during the upward motion of the ball? (b) What are the velocity and acceleration of the ball at the highest point of its motion? (c) Choose the = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of -ais, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion. (d) To what height does the ball rise and after how long does the ball return to the player s hands? (Take g = 9.8 m s and neglect air resistance).

MOTION IN A STRAIGHT LINE 57 3. Read each statement below carefully and state with reasons and eamples, if it is true or false ; A particle in one-dimensional motion (a) with zero speed at an instant may have non-zero acceleration at that instant (b) with zero speed may have non-zero velocity, (c) with constant speed must have zero acceleration, (d) with positive value of acceleration must be speeding up. 3. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to s. 3.3 Eplain clearly, with eamples, the distinction between : (a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval; (b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true? [For simplicity, consider one-dimensional motion only]. 3.4 A man walks on a straight road from his home to a market.5 km away with a speed of 5 km h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this eercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!] 3.5 In Eercises 3.3 and 3.4, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why? 3.6 Look at the graphs (a) to (d) (Fig. 3.0) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle. Fig. 3.0

58 PHYSICS 3.7 Figure 3. shows the -t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t >0? If not, suggest a suitable physical contet for this graph. 3.8 A police van moving on a highway with a speed of 30 km h fires a bullet at a thief s car speeding away in the same direction with a speed of 9 km h. If the muzzle speed of the bullet is 50 m s, with what speed does the bullet hit the thief s car? (Note: Obtain that speed which is relevant for damaging the thief s car). 3.9 Suggest a suitable physical situation for each of the following graphs (Fig 3.): Fig. 3. Fig. 3. 3.0 Figure 3.3 gives the -t plot of a particle eecuting one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter4). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s,. s,. s. Fig. 3.3 3. Figure 3.4 gives the -t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval. Fig. 3.4

MOTION IN A STRAIGHT LINE 59 3. Figure 3.5 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D? Fig. 3.5 Additional Eercises 3.3 A three-wheeler starts from rest, accelerates uniformly with m s on a straight road for 0 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the n th second (n =,,3.) versus n. What do you epect this plot to be during accelerated motion : a straight line or a parabola? 3.4 A boy standing on a stationary lift (open from above) throws a ball upwards with the maimum initial speed he can, equal to 49 m s. How much time does the ball take to return to his hands? If the lift starts moving up with a uniform speed of 5 m s - and the boy again throws the ball up with the maimum speed he can, how long does the ball take to return to his hands? 3.5 On a long horizontally moving belt (Fig. 3.6), a child runs to and fro with a speed 9 km h (with respect to the belt) between his father and mother located 50 m apart on the moving belt. The belt moves with a speed of 4 km h. For an observer on a stationary platform outside, what is the (a) speed of the child running in the direction of motion of the belt?. (b) speed of the child running opposite to the direction of motion of the belt? (c) time taken by the child in (a) and (b)? Which of the answers alter if motion is viewed by one of the parents? Fig. 3.6 3.6 Two stones are thrown up simultaneously from the edge of a cliff 00 m high with initial speeds of 5 m s and 30 m s. Verify that the graph shown in Fig. 3.7 correctly represents the time variation of the relative position of the second stone with respect to the first. Neglect air resistance and assume that the stones do not rebound after hitting the ground. Take g = 0 m s. Give the equations for the linear and curved parts of the plot.